HumptyDumpty wrote:
Could someone please explain me why this is illegal?:
\(\sqrt{x}+\sqrt{y}>0\)
\(\sqrt{x}>-\sqrt{y}\)
\((\sqrt{x})^2>(-\sqrt{y})^2\)
\(x>y\)
While this is legal:
\(\sqrt{x}-\sqrt{y}>0\)
\(\sqrt{x}>\sqrt{y}\)
\((\sqrt{x})^2>(\sqrt{y})^2\)
\(x>y\)
Thank you.
Writing \(x>y\) from \(\sqrt{x}+\sqrt{y}>0\) is not right. We have that the sum of two non-negative values (\(\sqrt{x}\) and \(\sqrt{y}\)) is greater than zero. How can we know that \(x>y\) from that? With the same logic you could get that \(y>x\). Right?
Algebraic explanation:
we can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);
But consider the case when one side is negative: \(-2<1\) --> if we square we get \(4<1\), which is not right. So, squaring an inequality where one side is negative won't always give the correct result.
In the first case \(-\sqrt{y}\leq{0}\), thus we cannot apply squaring. While in the second case both parts of the inequality are non-negatve, thus we can safely square.
GENERAL RULE:
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);
But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.
B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).
Hope it helps.
P.S. Can you please post the question from which you took that example? Thank you.
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